Entropy of spin-1/2 Paramagnetic gas

[curious_mind]

Homework Statement

Find the Entropy of spin-1/2 Paramagnetic gas

Relevant Equations

$S = k_B \ln \Omega(U)$
$S = \Sigma_r P_r \ln P_r$.

As we know, dipole can be only arranged either parallel or anti-parallel with respect to applied magnetic field $\vec{H}$ if we are to use quantum mechanical description, then parallel magnetic dipoles will have energy $\mu H$ and anti-parallel magnetic dipoles have energy $-\mu H$.
Method 1:
Let the number of magnetic dipoles with $m_s = + \frac{1}{2}$ spin are $n_{\uparrow}$ and number of magnetic dipoles with $m_s = - \frac{1}{2}$ spin are $n_{\downarrow}$. Total number of dipoles are $N = n_{\uparrow} + n_{\downarrow}$. So, $n_{\downarrow} = N - n_{\uparrow}$. And, total internal energy $U = n_{\uparrow} \mu H - n_{\downarrow} \mu H = ( 2 n_{\uparrow} - N ) \mu H \Longrightarrow n_{\uparrow} = \dfrac{1}{2} \left( \dfrac{U}{\mu H} + N \right)$.

Now, total number of microstates having internal energy U will be $$\Omega (U) = \dfrac{N!}{n_{\uparrow} ! ( N - n_{\uparrow} ) !}$$. Using $S = k_B ln \Omega(U)$, we get $S = k_B [ \ln N! - \ln n_{\uparrow} ! - \ln (N - n_{\uparrow}) ! ]$. Using stirling approximation, $$S = k_B [ N ln N - N - (n_{\uparrow} \ln n_{\uparrow} - n_{\uparrow}) - ( (N - n_{\uparrow}) \ln (N - n_{\uparrow}) - (N - n_{\uparrow}) ) ] $$. Upon simplifying this we get $$S = k_B \left[ N \ln N - \left( \dfrac{N \mu H- U }{2 \mu H} \right) \ln \left( \dfrac{N \mu H- U }{2 \mu H} \right) - \left( \dfrac{N \mu H+ U }{2 \mu H} \right) \ln \left( \dfrac{N \mu H+ U }{2 \mu H}\right) \right]$$.

Method 2:
Using expression $S = N k_B \Sigma_r P_r ln P_r$, we have $S = N k_B [ p \ln (p) + (1-p) \ln (1-p) ]$. If we take thermodynamic probability $p = \dfrac{e^{\frac{- \mu H}{k_B T}}}{e^{\frac{- \mu H}{k_B T}} + e^{\frac{\mu H}{k_B T}}}$. Upon simplifying this, I do not seem to get above expression. Answer comes in terms of hyperbolic functions.

Now, my confusion is - which expression is correct one - method 1 or method 2 ? Answer is given - $$S = N k_B \left[\left( \dfrac{-N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H+U }{2 \mu H} \right) - \left( \dfrac{N \mu H+ U }{2 N \mu H} \right) \ln \left( \dfrac{N \mu H- U }{2 N \mu H}\right) \right]$$. . Now, I am not getting this expression from above ones. This is nearer from first method, but it does not seem obtainable from first method as well. I think Answer can be incorrect.I am confused about which formula to use. Of course, only one answer is there. But I require clarification about which step I can proceed with under what Conditions.I am beginner in this subject. So I want clarification like, in what limit does this stirling approximation apply ? When temperature tend toward infinity or something else ? I have seen expression $S = N k_B \ln 2$ for infinite temperature limit. and $S = 0$ for $T \rightarrow 0$.

Kindly share any relevant link or resource which address this issue. I am looking into Statistical Mechanics book by F.Rief, it has section in 6th chapter for using approximations for canonical ensembles. It describes that calculation of number of microstates $\Omega(E)$ in range $E < E_r < E + \delta E$ can be difficult to calculate. Is this related to this? I don't think but still mentioning in case I am missing out something.

Thanks.

 

[yezia]

I did not read everything, but a quick suggestion for method 1: try to express the whole thing in terms of $\dfrac{U}{\mu H}$ and factor out (it's normal if some of your factorizations won't lead anywhere) $N$. I don't know the exact expression of the entropy, and since you have doubt on the solution given, you might want instead to compare with the internal energy which has a very simple expression:
$$
U = -N\mu H\tanh\left(\dfrac{\mu H}{k_BT}\right).
$$
First try my suggestion, then compute $\beta :=\dfrac{1}{k_BT} = \dfrac{\partial \ln\Omega(U)}{\partial U}$, and finally express $U$ in terms of everything else. One useful relation: $\mathrm{argtanh}\left(x\right)=\dfrac{1}{2}\ln\left(\dfrac{1+x}{1-x}\right).$

First method uses the microcanonical ensemble while second uses the canonical ensemble, that's the main difference ... For a system of spins, we usually use the first method when we ... don't know yet the canonical ensemble: second method is actually much faster. I'm pretty sure your coursebook (in particular Reif) explains it very well.

 

[TSny]

I'll add a couple of more comments. Your answer for $S$ as a function of $U$ looks ok to me.

See what your result gives for $S$ in the limit $T = 0$ and the limit $T = \infty$. For $T = 0$ what do you expect for the value of $U$? What does this value of $U$ give for $S$? Repeat for $T = \infty$.

What does their result for $S$ give in these limits?

 

[curious_mind]

I'll add a couple of more comments. Your answer for $S$ as a function of $U$ looks ok to me.

See what your result gives for $S$ in the limit $T = 0$ and the limit $T = \infty$. For $T = 0$ what do you expect for the value of $U$? What does this value of $U$ give for $S$? Repeat for $T = \infty$.

What does their result for $S$ give in these limits?

At $T=0$ I expect $U=0$, is it correct ? And using $U=0$ and expression of method 1 of microcanonical ensemble(as pointed out by @yezia ) I am getting $S = N k_B \ln 2$ .

Ok then i wrote in the question, as $T \rightarrow 0$ , $S =0$ would not be correct then.

For $T \rightarrow \infty$ , I am not sure about behavior of $U$. I am still confused about this. I lack some basics.

In microcanonical ensemble, $U$ is fixed, right ? So expressions of microcanonical ensemble will have $U$ in it and thermodynamic functions of canonical ensemble will have $T$ in it. This thing I missed totally.

Yes, these things explained in book. But often books are large and you miss lot of things despite reading them.

 

[yezia]

When $T\rightarrow 0$, the internal energy is not 0. $T = 0$ corresponds to the state of minimal energy, and in this case, all the spins must be anti-parallel to the magnetic field (do you see why ?). What is the expression of $U(T=0)$ now that you know this ?

On the other hand, when $T\rightarrow \infty$ - important: you cannot physically say that a variable $x$ approaches infinity, we usually introduce characteristic variables $x^*$ (it is all a matter of dimensions) and judge "big" or "small" comparatively to these -, you can consider each spin as totally "indepedent"; they do as they want, and they have same probability of being up or down (do you understand why ?). Knowing this, what would $U(T\gg T^*)$ compute to ?

Other tip: don't forget the expression $S\propto \ln\Omega$. At 0 temperature, how many states are accessible by the system ? At very high temperature, each spin has 2 equiprobable state; how many states are therefore accessible by the whole system ? This can help you understand the value of the entropy at the mentioned limits.

You can moreover look at the graph of $U=U(T)$ (expression in my first message) to enhance your intuition and understand better the asymptotic behaviour of $U(T)$.

In microcanonical ensemble, $U$ is fixed, right ? So expressions of microcanonical ensemble will have $U$ in it and thermodynamic functions of canonical ensemble will have $T$ in it. This thing I missed totally.

Yes, these things explained in book. But often books are large and you miss lot of things despite reading them.

That's pretty much it yes! It is perfectly normal to not understand everything at first when reading a big book as Reif's :)

 

[DrClaude]

I am confused about which formula to use. Of course, only one answer is there. But I require clarification about which step I can proceed with under what Conditions.

The first formula is a function of $U$, so it works for a fixed energy. The second, as you wrote yourself, involves temperature, so it works for fixed $T$. Which to use depends on the particular situation at hand.

in what limit does this stirling approximation apply ?

It works when $N$ is big. The approximation is better than 1% already for $N=100$.

 

[curious_mind]

When $T\rightarrow 0$, the internal energy is not 0. $T = 0$ corresponds to the state of minimal energy, and in this case, all the spins must be anti-parallel to the magnetic field (do you see why ?). What is the expression of $U(T=0)$ now that you know this ?

On the other hand, when $T\rightarrow \infty$ - important: you cannot physically say that a variable $x$ approaches infinity, we usually introduce characteristic variables $x^*$ (it is all a matter of dimensions) and judge "big" or "small" comparatively to these -, you can consider each spin as totally "indepedent"; they do as they want, and they have same probability of being up or down (do you understand why ?). Knowing this, what would $U(T\gg T^*)$ compute to ?

Other tip: don't forget the expression $S\propto \ln\Omega$. At 0 temperature, how many states are accessible by the system ? At very high temperature, each spin has 2 equiprobable state; how many states are therefore accessible by the whole system ? This can help you understand the value of the entropy at the mentioned limits.

You can moreover look at the graph of $U=U(T)$ (expression in my first message) to enhance your intuition and understand better the asymptotic behaviour of $U(T)$.That's pretty much it yes! It is perfectly normal to not understand everything at first when reading a big book as Reif's :)

Ok, so , since minimum energy will take place at T=0, $U = - N \mu B H$ that's why all diploes will align anti-parallely. For $T \rightarrow \infty$ I think I understood from your description and your previous first reply.

Thanks. :)

 

[Vineyo]

According to what I learned from this article (page 23-24), the first one is an approximation of the second one, when $\Omega(E)$ increases rapidly and $E_*$ gives the maximum value of $\Omega(E)e^{-βE}$. So the partition function $Z≈\Omega(E_*)e^{-βE_*}$, thus $S≈k_Bln\Omega(E_*)$.